Use the Ratio Test to determine whether each series converges or diverges.

Solution (a) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{k}{4^{k}}$ is a series of nonzero terms; $a_{n+1}=\dfrac{n+1}{4^{n+1}}$ and $a_{n}= \dfrac{n}{4^{n}}$. The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\dfrac{\dfrac{n+1}{4^{n+1}}}{ \dfrac{n}{4^{n}}}=\dfrac{n+1}{4^{n+1}}\cdot \dfrac{4^{n}}{n}=\frac{n+1}{4n}$$

Then $$\lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{n+1}{4n}=\frac{1}{4} <1$$

Since the limit is less than $1$, the series converges.

(b) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{2^{k}}{k}$ is a series of nonzero terms; $a_{n+1}=\dfrac{2^{n+1}}{n+1}$ and $a_{n}=\dfrac{ 2^{n}}{n}.$ The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\frac{2^{n+1}}{n+1} \cdot \frac{ n}{2^{n}} =\frac{2n}{n+1}$$

Then $$\lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{2n}{n+1}=2>1$$

Since the limit is greater than $1$, the series diverges.

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(c) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{3k+1}{k^{2}}$ is a series of nonzero terms; $a_{n+1}=\dfrac{3(n+1) +1}{(n+1) ^{2}}=\dfrac{3n+4}{(n+1) ^{2}}$ and $a_{n}=\dfrac{3n+1 }{n^{2}}$. The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert {=}\left[ \frac{3n+4}{(n+1)^{2}} \right] \left( \frac{n^{2}}{3n+1}\right) {=}\left( \frac{3n+4}{3n+1}\right) \left( \frac{n^{2}}{n^{2}+2n+1}\right) =\dfrac{3n^{3}+4n^{2}}{3n^{3}+7n^{2}+5n+1}$$

Then $$\lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\dfrac{3n^{3}+4n^{2}}{3n^{3}+7n^{2}+5n+1}=1$$

The Ratio Test gives no information about this series. Another test must be used. $\bigg({\rm{You}}\, {\rm{can}}\, {\rm{show}}\, {\rm{that}}\, {\rm{the}}\, {\rm{series}}\, {\rm{diverges}}\, {\rm{by}}\, {\rm{comparing}}\, {\rm{it}}\, {\rm{to}}\, {\rm{the}}\, {\rm{harmonic}}\, {\rm{series}}\, \sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}.\bigg)$

(d) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}$ is a series of nonzero terms; $a_{n+1}=\dfrac{1}{(n+1) !}$ and $a_{n}=\dfrac{ 1}{n!}$. The absolute value of their ratio is $$\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\frac{n!}{(n+1)!} = \frac{1}{n+1}$$

Then $$\lim_{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim_{n\rightarrow \infty }\frac{1}{n+1}=0$$

Since the limit is less than $1,$ the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}$ converges.